A container consists mixture of four gases as $5 \, \mathrm{g} \, \mathrm{H}_2$, $8 \, \mathrm{g} \,…

A container consists mixture of four gases as $5 \, \mathrm{g} \, \mathrm{H}_2$, $8 \, \mathrm{g} \, \mathrm{He}$, $50 \, \mathrm{g} \, \mathrm{CO}_2$ and $20 \, \mathrm{g} \, \mathrm{Ne}$ at a certain temperature. Which of the following gases exerts minimum partial pressure? $\begin{aligned} \begin{array}{|c|c|c|c|c|} \hline X=x & 1 & 2 & 3 & 4 \\ \hline P(X=x) & 0.1 & 0.2 & 0.3 & 0.4 \\ \hline \end{array} \end{aligned}$
  1. $\mathrm{H}_2$
  2. He
  3. $\mathrm{CO}_2$
  4. Ne

Solution

Dalton's Law of Partial Pressures states that the partial pressure of each gas in a mixture is the product of its mole fraction and the total pressure: $P_i = X_i P_{\text{total}}$. Since $P_{\text{total}}$ is constant, the partial pressure is proportional to the mole fraction, which depends on the number of moles: $X_i = \frac{n_i}{n_{\text{total}}}$. Therefore, the gas with the fewest moles exerts the minimum partial pressure.

Moles are determined from mass and molar mass using $n = \frac{\text{mass}}{\text{molar mass}}$.
The molar masses are: $2$ g/mol for $\mathrm{H}_2$, $4$ g/mol for $\mathrm{He}$, $44$ g/mol for $\mathrm{CO}_2$, and $20$ g/mol for $\mathrm{Ne}$.

Calculating for each gas:
$n_{\mathrm{H}_2} = \frac{5}{2} = 2.5$ mol
$n_{\mathrm{He}} = \frac{8}{4} = 2.0$ mol
$n_{\mathrm{CO}_2} = \frac{50}{44} \approx 1.136$ mol
$n_{\mathrm{Ne}} = \frac{20}{20} = 1.0$ mol

The minimum number of moles is $1.0$ mol for neon, which will exert the minimum partial pressure.

$\boxed{\text{D}}$

Asked in: MHT CET 2025 (26 April Shift 2)

Practice more States of Matter questions on Aicharya